Confusing Question on "Degrees of Freedom"

Question

I am studying “Rosenberg’s Analytical Dynamics of Discrete Systems”. In chapter 4, he discussed on holonomic and nonholonomic constraints. At the end of the chapter, he asked a question that confused me:

“Particle P can move on the bottom of a two-dimensional cage. Since the position of the particle can be described by the two coordinates x and y, or by the three coordinates $\theta$, $\phi$ and $\xi$, it seems that the latter must satisfy a constraint relation between them. Can you find that relation? How many degrees of freedom does the particle have?”

If we use $\theta$, $\phi$ and $\xi$ to describe the position of P, it means P has three degree of freedom. In the contrary, because of planar motion of the particle P, there are utmost two degree of freedom. It seems paradoxical. What is my mistake?

Answer 1

The mistake is just here

If we use θ, ϕ and ξ to describe the position of P, it means P has three degree of freedom.

The number of degrees of freedom is not the number of variables you use to describe a system (let's take it very general), but the number of independent variables needed.

Indeed, in text you provided the author says

Since the position of the particle can be described by the two coordinates x and y, or by the three coordinates θ, ϕ and ξ, it seems that the latter must satisfy a constraint relation between them

which can be rephrased as follows: since we need only two variables to fully describe the position ($x$ and $y$, so two degrees of freedom), if we want to use three angular variable, then they cannot be all independent and there must be a relation between them (so that the value of one of the angle is determined by the other two).

Edit Let me add a quick example. Suppose you want to describe the motion of a child on a carousel. Being in a three-dimensional space, the position of the child would be described by three coordinates $x$, $y$ and $z$. However, the number of degrees of freedom here is just 1, because you only need one variable to describe the motion (for example the angle with respect to the center of the carousel).

Answer 2

In two dimensions, without constraints, the cage would have 3 degrees of freedom (we can take as its generalized coordinates the two Cartesian coordinates $X_h, Y_h$ of the hanging point, and the angle $\phi$ that determines its orientation in the plane). On the other hand, the particle would have 2 DOF (we can take its $x$ and $y$ coordinates as the generalized ones). So, the system would have five DOF without constraints.

However, this system has two constraints: i) the hanging point is at a constant distance $l$ from the origin $O$ $$X_h^2+Y_h^2-l^2=0,$$ and ii) the particle moves on the bottom of the cage, that is (if my algebra is correct!) $$(x X_h + y Y_h) \cos\phi + (y X_h-x Y_h)\sin\phi = l (b+l\cos\phi).$$ These are holonomic constraints. So, after using the constraint relations, the system has three degrees of freedom. We can write $X_h$ and $Y_h$ in terms of the angle $\theta:$
$$X_h = l \sin\theta,\;Y_h=l\cos\theta,$$ so we can say that the cage has two DOF (angles $\theta$ and $\phi$), while the particle has one degree of freedom (coordinate $\xi$), $$x = l \sin\theta + b\sin(\theta-\phi) +\xi \cos(\theta-\phi),\;\; y = l\cos\theta + b\cos(\theta-\phi) - \xi \sin(\theta-\phi). $$ Of course, as the coordinate $\xi$ is determined with respect to the cage, $x$ and $y$ depend on the cage angles.

A simpler way of thinking: the particle is moving along a curve, so it has only one degree of freedom. Perhaps, it is striking that the curve is not fixed, as it depends on the cage motion, which in turn is affected by the particle motion.

Answer 3

I don't know whether Rosenberg has their own definition degrees of freedom of a particle, but the system clearly has three DoF (if planar motion is assumed). Then, the two Cartesian coordinates of the particle can be expressed through the three generalized coordinates of the system.

Saying that the particle moves in a plane completely ignores the fact that the particle is not the only part of the system moving (if it was, it would have just one DoF, anyway, because it can't leave the bottom of the – per assumption immovable – cage). There's the suspension point of the cage and its rotation about it as well, and as such the configuration space is of higher dimension than $\mathbb{R}^2$. Indeed, we know with these constraints it is a three-dimensional manifold of some sort.